Question

logbase10(x²-3x+6)=1

Answer 1

ANSWER

ANSWERGiven: log

ANSWERGiven: log 10

ANSWERGiven: log 10

ANSWERGiven: log 10 (x

ANSWERGiven: log 10 (x 2

ANSWERGiven: log 10 (x 2 −3x+6)=1

ANSWERGiven: log 10 (x 2 −3x+6)=1log

ANSWERGiven: log 10 (x 2 −3x+6)=1log 10

ANSWERGiven: log 10 (x 2 −3x+6)=1log 10

ANSWERGiven: log 10 (x 2 −3x+6)=1log 10 (x

ANSWERGiven: log 10 (x 2 −3x+6)=1log 10 (x 2

ANSWERGiven: log 10 (x 2 −3x+6)=1log 10 (x 2 −3x+6)=log

ANSWERGiven: log 10 (x 2 −3x+6)=1log 10 (x 2 −3x+6)=log 10

ANSWERGiven: log 10 (x 2 −3x+6)=1log 10 (x 2 −3x+6)=log 10

ANSWERGiven: log 10 (x 2 −3x+6)=1log 10 (x 2 −3x+6)=log 10 10.......(1) [since log

ANSWERGiven: log 10 (x 2 −3x+6)=1log 10 (x 2 −3x+6)=log 10 10.......(1) [since log a

ANSWERGiven: log 10 (x 2 −3x+6)=1log 10 (x 2 −3x+6)=log 10 10.......(1) [since log a

ANSWERGiven: log 10 (x 2 −3x+6)=1log 10 (x 2 −3x+6)=log 10 10.......(1) [since log a a=1]

From (1):

From (1):x

From (1):x 2

From (1):x 2 −3x+6=10

From (1):x 2 −3x+6=10⇒x

From (1):x 2 −3x+6=10⇒x 2

From (1):x 2 −3x+6=10⇒x 2 −3x−4=0

From (1):x 2 −3x+6=10⇒x 2 −3x−4=0⇒(x−4)(x+1)=0

From (1):x 2 −3x+6=10⇒x 2 −3x−4=0⇒(x−4)(x+1)=0⇒x=−1 or 4

From (1):x 2 −3x+6=10⇒x 2 −3x−4=0⇒(x−4)(x+1)=0⇒x=−1 or 4But we should also check that f(x)=x

From (1):x 2 −3x+6=10⇒x 2 −3x−4=0⇒(x−4)(x+1)=0⇒x=−1 or 4But we should also check that f(x)=x 2

From (1):x 2 −3x+6=10⇒x 2 −3x−4=0⇒(x−4)(x+1)=0⇒x=−1 or 4But we should also check that f(x)=x 2 −3x+6>0 for x=−1,4........[for

From (1):x 2 −3x+6=10⇒x 2 −3x−4=0⇒(x−4)(x+1)=0⇒x=−1 or 4But we should also check that f(x)=x 2 −3x+6>0 for x=−1,4........[for log

From (1):x 2 −3x+6=10⇒x 2 −3x−4=0⇒(x−4)(x+1)=0⇒x=−1 or 4But we should also check that f(x)=x 2 −3x+6>0 for x=−1,4........[for log a

From (1):x 2 −3x+6=10⇒x 2 −3x−4=0⇒(x−4)(x+1)=0⇒x=−1 or 4But we should also check that f(x)=x 2 −3x+6>0 for x=−1,4........[for log a

From (1):x 2 −3x+6=10⇒x 2 −3x−4=0⇒(x−4)(x+1)=0⇒x=−1 or 4But we should also check that f(x)=x 2 −3x+6>0 for x=−1,4........[for log a N; N>0]

From (1):x 2 −3x+6=10⇒x 2 −3x−4=0⇒(x−4)(x+1)=0⇒x=−1 or 4But we should also check that f(x)=x 2 −3x+6>0 for x=−1,4........[for log a N; N>0]f(−1)=(−1)

From (1):x 2 −3x+6=10⇒x 2 −3x−4=0⇒(x−4)(x+1)=0⇒x=−1 or 4But we should also check that f(x)=x 2 −3x+6>0 for x=−1,4........[for log a N; N>0]f(−1)=(−1) 2

From (1):x 2 −3x+6=10⇒x 2 −3x−4=0⇒(x−4)(x+1)=0⇒x=−1 or 4But we should also check that f(x)=x 2 −3x+6>0 for x=−1,4........[for log a N; N>0]f(−1)=(−1) 2 −3(−1)+6=10>0

From (1):x 2 −3x+6=10⇒x 2 −3x−4=0⇒(x−4)(x+1)=0⇒x=−1 or 4But we should also check that f(x)=x 2 −3x+6>0 for x=−1,4........[for log a N; N>0]f(−1)=(−1) 2 −3(−1)+6=10>0f(4)=4

From (1):x 2 −3x+6=10⇒x 2 −3x−4=0⇒(x−4)(x+1)=0⇒x=−1 or 4But we should also check that f(x)=x 2 −3x+6>0 for x=−1,4........[for log a N; N>0]f(−1)=(−1) 2 −3(−1)+6=10>0f(4)=4 2

From (1):x 2 −3x+6=10⇒x 2 −3x−4=0⇒(x−4)(x+1)=0⇒x=−1 or 4But we should also check that f(x)=x 2 −3x+6>0 for x=−1,4........[for log a N; N>0]f(−1)=(−1) 2 −3(−1)+6=10>0f(4)=4 2 −3(4)+6=10>0

From (1):x 2 −3x+6=10⇒x 2 −3x−4=0⇒(x−4)(x+1)=0⇒x=−1 or 4But we should also check that f(x)=x 2 −3x+6>0 for x=−1,4........[for log a N; N>0]f(−1)=(−1) 2 −3(−1)+6=10>0f(4)=4 2 −3(4)+6=10>0Hence both values are satisfying the condition.

From (1):x 2 −3x+6=10⇒x 2 −3x−4=0⇒(x−4)(x+1)=0⇒x=−1 or 4But we should also check that f(x)=x 2 −3x+6>0 for x=−1,4........[for log a N; N>0]f(−1)=(−1) 2 −3(−1)+6=10>0f(4)=4 2 −3(4)+6=10>0Hence both values are satisfying the condition.So, x=−1,4

Answer 2

Step-by-step explanation:

Given: lg10(x ^2 −3x+6)=1

log 10 (x^ 2 −3x+6)=log10.......(1)

[since log a=1]

From (1):

x^2 −3x+6=10

⇒x°2 −3x−4=0

⇒(x−4)(x+1)=0

⇒x=−1 or 4

But we should also check that f(x)=x^2 −3x+6>0 for x=−1,4........[for

log a N; N>0]

f(−1)=(−1)^2 −3(−1)+6=10>0

f(4)=(4)^2 −3(4)+6=10>0

Hence both values are satisfying the condition.

So, x=−1,4